find-I-a-b-0-sin-ax-sin-bx-x-2-dx-witha-gt-0-and-b-gt-0-

Question Number 79094 by mathmax by abdo last updated on 22/Jan/20

f i n d I_{a, b} = \int_{0}^{\infty} \frac{s i n (a x) s i n (b x)}{x^{2}} d x w i t h a > 0 a n d b > 0

Answered by mind is power last updated on 23/Jan/20

2 s i n (a x) s i n (b x) = c o s ((a - b) x) - c o s ((a + b) x)

f (z) = \int_{0}^{+ \infty} \frac{c o s (z x) - 1}{x^{2}} d x

z \in R w e l l d e f i n d

i n z e r o c o s (z x) - 1 = - \frac{z^{2} x^{2}}{2} + o (x^{3})

f o r z > 1 \frac{∣ c o s (z x) - 1 ∣}{x^{2}} ⩽ \frac{2}{x^{2}}, i n t e g r a b l i n [1, + \infty [

\forall (z, x) \in I R \times R^{*} (z, x) \to \frac{c o s (z x) - 1}{x^{2}} \in C_{\infty} C_{1} i s w h a t w e n e e d

\frac{\partial}{\partial z} (\frac{c o s (z x) - 1}{x^{2}}) = - \frac{s i n (z x)}{x} \in C_{1}] 0, + \infty [

\int_{0}^{+ \infty} - \frac{s i n (z x)}{x} d x < \infty \Rightarrow f (z) \in C_{1}

f^{'} (z) = \int_{0}^{+ \infty} \frac{\partial}{\partial z} (\frac{c o s (z x) - 1}{x^{2}}) = - \int_{0}^{+ \infty} \frac{s i n (z x)}{x} d x, u = z x

i f z > 0

= - \int_{0}^{+ \infty} \frac{s i n (u)}{u} d u = - \frac{π}{2}, z < 0 = \frac{π}{2}

\Rightarrow f^{'} (z) = - s i g n (z) \frac{π}{2}

f (z) = - \frac{z s i g n (z) π}{2} + c

f (0) = 0 \Rightarrow c = 0

f (z) = \frac{- z s i g n (z) π}{2}

I_{a, b} = \int_{0}^{+ \infty} \frac{s i n (a x) s i n (b x)}{x^{2}} d x

= \int_{0}^{+ \infty} \frac{c o s ((a - b) x) - c o s ((a + b) x)}{2 x^{2}} d x

= \int_{0}^{+ \infty} \frac{c o s ((a - b) x) - 1 - (c o s ((a + b) x) - 1)}{2 x^{2}} d x

= \frac{1}{2} \int_{0}^{+ \infty} \frac{c o s ((a - b) x) - 1}{x^{2}} - \frac{1}{2} \int_{0}^{+ \infty} \frac{c o s ((a + b) x) - 1}{x^{2}}

= \frac{1}{2} f ((a - b)) - \frac{1}{2} f (a + b)) = - \frac{(a - b) s i v n (a - b) π}{4} + \frac{(a + b) s i g n (a + b) π}{4}

= \frac{π}{4} ((a + b) s i g n (a + b) - (a - b) s i g n (a - b))

Commented by msup trace by abdo last updated on 23/Jan/20

t h a n k y o u s i r .